# Find the Number Occurring Odd Number of Times

• Difficulty Level : Easy
• Last Updated : 21 Nov, 2022

Given an array of positive integers. All numbers occur an even number of times except one number which occurs an odd number of times. Find the number in O(n) time & constant space.

Examples :

Input : arr = {1, 2, 3, 2, 3, 1, 3}
Output : 3

Input : arr = {5, 7, 2, 7, 5, 2, 5}
Output : 5

Recommended Practice

A Simple Solution is to run two nested loops. The outer loop picks all elements one by one and the inner loop counts the number of occurrences of the element picked by the outer loop. The time complexity of this solution is O(n2).

Below is the implementation of the brute force approach :

## C++

 `// C++ program to find the element``// occurring odd number of times``#include``using` `namespace` `std;` `// Function to find the element``// occurring odd number of times``int` `getOddOccurrence(``int` `arr[], ``int` `arr_size)``{``    ``for` `(``int` `i = 0; i < arr_size; i++) {``        ` `        ``int` `count = 0;``        ` `        ``for` `(``int` `j = 0; j < arr_size; j++)``        ``{``            ``if` `(arr[i] == arr[j])``                ``count++;``        ``}``        ``if` `(count % 2 != 0)``            ``return` `arr[i];``    ``}``    ``return` `-1;``}` `// driver code``int` `main()``    ``{``        ``int` `arr[] = { 2, 3, 5, 4, 5, 2,``                      ``4, 3, 5, 2, 4, 4, 2 };``        ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `        ``// Function calling``        ``cout << getOddOccurrence(arr, n);` `        ``return` `0;``    ``}`

## Java

 `// Java program to find the element occurring``// odd number of times``class` `OddOccurrence {``    ` `    ``// function to find the element occurring odd``    ``// number of times``    ``static` `int` `getOddOccurrence(``int` `arr[], ``int` `arr_size)``    ``{``        ``int` `i;``        ``for` `(i = ``0``; i < arr_size; i++) {``            ``int` `count = ``0``;``            ``for` `(``int` `j = ``0``; j < arr_size; j++) {``                ``if` `(arr[i] == arr[j])``                    ``count++;``            ``}``            ``if` `(count % ``2` `!= ``0``)``                ``return` `arr[i];``        ``}``        ``return` `-``1``;``    ``}``    ` `    ``// driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = ``new` `int``[]{ ``2``, ``3``, ``5``, ``4``, ``5``, ``2``, ``4``, ``3``, ``5``, ``2``, ``4``, ``4``, ``2` `};``        ``int` `n = arr.length;``        ``System.out.println(getOddOccurrence(arr, n));``    ``}``}``// This code has been contributed by Kamal Rawal`

## Python3

 `# Python program to find the element occurring``# odd number of times``    ` `# function to find the element occurring odd``# number of times``def` `getOddOccurrence(arr, arr_size):``    ` `    ``for` `i ``in` `range``(``0``,arr_size):``        ``count ``=` `0``        ``for` `j ``in` `range``(``0``, arr_size):``            ``if` `arr[i] ``=``=` `arr[j]:``                ``count``+``=``1``            ` `        ``if` `(count ``%` `2` `!``=` `0``):``            ``return` `arr[i]``        ` `    ``return` `-``1``    ` `    ` `# driver code``arr ``=` `[``2``, ``3``, ``5``, ``4``, ``5``, ``2``, ``4``, ``3``, ``5``, ``2``, ``4``, ``4``, ``2` `]``n ``=` `len``(arr)``print``(getOddOccurrence(arr, n))` `# This code has been contributed by``# Smitha Dinesh Semwal`

## C#

 `// C# program to find the element``// occurring odd number of times``using` `System;` `class` `GFG``{``    ``// Function to find the element``    ``// occurring odd number of times``    ``static` `int` `getOddOccurrence(``int` `[]arr, ``int` `arr_size)``    ``{``        ``for` `(``int` `i = 0; i < arr_size; i++) {``            ``int` `count = 0;``            ` `            ``for` `(``int` `j = 0; j < arr_size; j++) {``                ``if` `(arr[i] == arr[j])``                    ``count++;``            ``}``            ``if` `(count % 2 != 0)``                ``return` `arr[i];``        ``}``        ``return` `-1;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };``        ``int` `n = arr.Length;``        ``Console.Write(getOddOccurrence(arr, n));``    ``}``}` `// This code is contributed by Sam007`

## PHP

 ``

## Javascript

 ``

Output :

`5`

Time Complexity: O(n^2)
Auxiliary Space: O(1)

A Better Solution is to use Hashing. Use array elements as a key and their counts as values. Create an empty hash table. One by one traverse the given array elements and store counts. The time complexity of this solution is O(n). But it requires extra space for hashing.

Program :

## C++

 `// C++ program to find the element ``// occurring odd number of times``#include ``using` `namespace` `std;` `// function to find the element``// occurring odd number of times``int` `getOddOccurrence(``int` `arr[],``int` `size)``{``    ` `    ``// Defining HashMap in C++``    ``unordered_map<``int``, ``int``> hash;` `    ``// Putting all elements into the HashMap``    ``for``(``int` `i = 0; i < size; i++)``    ``{``        ``hash[arr[i]]++;``    ``}``    ``// Iterate through HashMap to check an element``    ``// occurring odd number of times and return it``    ``for``(``auto` `i : hash)``    ``{``        ``if``(i.second % 2 != 0)``        ``{``            ``return` `i.first;``        ``}``    ``}``    ``return` `-1;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 3, 5, 4, 5, 2, 4,``                    ``3, 5, 2, 4, 4, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ` `    ``// Function calling``    ``cout << getOddOccurrence(arr, n);` `    ``return` `0;``}` `// This code is contributed by codeMan_d.`

## Java

 `// Java program to find the element occurring odd``// number of times``import` `java.io.*;``import` `java.util.HashMap;` `class` `OddOccurrence``{``    ``// function to find the element occurring odd``    ``// number of times``    ``static` `int` `getOddOccurrence(``int` `arr[], ``int` `n)``    ``{``        ``HashMap hmap = ``new` `HashMap<>();``        ` `        ``// Putting all elements into the HashMap``        ``for``(``int` `i = ``0``; i < n; i++)``        ``{``            ``if``(hmap.containsKey(arr[i]))``            ``{``                ``int` `val = hmap.get(arr[i]);``                        ` `                ``// If array element is already present then``                ``// increase the count of that element.``                ``hmap.put(arr[i], val + ``1``);``            ``}``            ``else``                ` `                ``// if array element is not present then put``                ``// element into the HashMap and initialize``                ``// the count to one.``                ``hmap.put(arr[i], ``1``);``        ``}` `        ``// Checking for odd occurrence of each element present``        ``// in the HashMap``        ``for``(Integer a:hmap.keySet())``        ``{``            ``if``(hmap.get(a) % ``2` `!= ``0``)``                ``return` `a;``        ``}``        ``return` `-``1``;``    ``}``        ` `    ``// driver code   ``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = ``new` `int``[]{``2``, ``3``, ``5``, ``4``, ``5``, ``2``, ``4``, ``3``, ``5``, ``2``, ``4``, ``4``, ``2``};``        ``int` `n = arr.length;``        ``System.out.println(getOddOccurrence(arr, n));``    ``}``}``// This code is contributed by Kamal Rawal`

## Python3

 `# Python3 program to find the element ``# occurring odd number of times`` ` `# function to find the element``# occurring odd number of times``def` `getOddOccurrence(arr,size):``     ` `    ``# Defining HashMap in C++``    ``Hash``=``dict``()`` ` `    ``# Putting all elements into the HashMap``    ``for` `i ``in` `range``(size):``        ``Hash``[arr[i]]``=``Hash``.get(arr[i],``0``) ``+` `1``;``    ` `    ``# Iterate through HashMap to check an element``    ``# occurring odd number of times and return it``    ``for` `i ``in` `Hash``:` `        ``if``(``Hash``[i]``%` `2` `!``=` `0``):``            ``return` `i``    ``return` `-``1` ` ` `# Driver code``arr``=``[``2``, ``3``, ``5``, ``4``, ``5``, ``2``, ``4``,``3``, ``5``, ``2``, ``4``, ``4``, ``2``]``n ``=` `len``(arr)`` ` `# Function calling``print``(getOddOccurrence(arr, n))` `# This code is contributed by mohit kumar`

## C#

 `// C# program to find the element occurring odd``// number of times``using` `System;``using` `System.Collections.Generic;` `public` `class` `OddOccurrence``{``    ``// function to find the element occurring odd``    ``// number of times``    ``static` `int` `getOddOccurrence(``int` `[]arr, ``int` `n)``    ``{``        ``Dictionary<``int``,``int``> hmap = ``new` `Dictionary<``int``,``int``>();``        ` `        ``// Putting all elements into the HashMap``        ``for``(``int` `i = 0; i < n; i++)``        ``{``            ``if``(hmap.ContainsKey(arr[i]))``            ``{``                ``int` `val = hmap[arr[i]];``                        ` `                ``// If array element is already present then``                ``// increase the count of that element.``                ``hmap.Remove(arr[i]);``                ``hmap.Add(arr[i], val + 1);``            ``}``            ``else``                ` `                ``// if array element is not present then put``                ``// element into the HashMap and initialize``                ``// the count to one.``                ``hmap.Add(arr[i], 1);``        ``}` `        ``// Checking for odd occurrence of each element present``        ``// in the HashMap``        ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `hmap)``        ``{``            ``if``(entry.Value % 2 != 0)``            ``{``                ``return` `entry.Key;``            ``}``        ``}``        ``return` `-1;``    ``}``        ` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `[]arr = ``new` `int``[]{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};``        ``int` `n = arr.Length;``        ``Console.WriteLine(getOddOccurrence(arr, n));``    ``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``

Output :

`5`

Time Complexity: O(n)
Auxiliary Space: O(n)

The Best Solution is to do bitwise XOR of all the elements. XOR of all elements gives us odd occurring elements.

```Here ^ is the XOR operators;
Note :
x^0 = x
x^y=y^x (Commutative property holds)
(x^y)^z = x^(y^z) (Distributive property holds)
x^x=0```

Below is the implementation of the above approach.

## C++

 `// C++ program to find the element``// occurring odd number of times``#include ``using` `namespace` `std;` `// Function to find element occurring``// odd number of times``int` `getOddOccurrence(``int` `ar[], ``int` `ar_size)``{``    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < ar_size; i++)    ``        ``res = res ^ ar[i];``    ` `    ``return` `res;``}` `/* Driver function to test above function */``int` `main()``{``    ``int` `ar[] = {2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};``    ``int` `n = ``sizeof``(ar)/``sizeof``(ar[0]);``    ` `    ``// Function calling``    ``cout << getOddOccurrence(ar, n);``    ` `    ``return` `0;``}`

## C

 `// C program to find the element``// occurring odd number of times``#include ` `// Function to find element occurring``// odd number of times``int` `getOddOccurrence(``int` `ar[], ``int` `ar_size)``{``    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < ar_size; i++)    ``        ``res = res ^ ar[i];``    ` `    ``return` `res;``}` `/* Driver function to test above function */``int` `main()``{``    ``int` `ar[] = {2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};``    ``int` `n = ``sizeof``(ar) / ``sizeof``(ar[0]);``    ` `    ``// Function calling``    ``printf``(``"%d"``, getOddOccurrence(ar, n));``    ``return` `0;``}`

## Java

 `//Java program to find the element occurring odd number of times` `class` `OddOccurance``{``    ``int` `getOddOccurrence(``int` `ar[], ``int` `ar_size)``    ``{``        ``int` `i;``        ``int` `res = ``0``;``        ``for` `(i = ``0``; i < ar_size; i++)``        ``{``            ``res = res ^ ar[i];``        ``}``        ``return` `res;``    ``}` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``OddOccurance occur = ``new` `OddOccurance();``        ``int` `ar[] = ``new` `int``[]{``2``, ``3``, ``5``, ``4``, ``5``, ``2``, ``4``, ``3``, ``5``, ``2``, ``4``, ``4``, ``2``};``        ``int` `n = ar.length;``        ``System.out.println(occur.getOddOccurrence(ar, n));``    ``}``}``// This code has been contributed by Mayank Jaiswal`

## Python3

 `# Python program to find the element occurring odd number of times` `def` `getOddOccurrence(arr):` `    ``# Initialize result``    ``res ``=` `0``    ` `    ``# Traverse the array``    ``for` `element ``in` `arr:``        ``# XOR with the result``        ``res ``=` `res ^ element` `    ``return` `res` `# Test array``arr ``=` `[ ``2``, ``3``, ``5``, ``4``, ``5``, ``2``, ``4``, ``3``, ``5``, ``2``, ``4``, ``4``, ``2``]` `print``(``"%d"` `%` `getOddOccurrence(arr))`

## C#

 `// C# program to find the element``// occurring odd number of times``using` `System;` `class` `GFG``{``    ``// Function to find the element``    ``// occurring odd number of times``    ``static` `int` `getOddOccurrence(``int` `[]arr, ``int` `arr_size)``    ``{``        ``int` `res = 0;``        ``for` `(``int` `i = 0; i < arr_size; i++)``        ``{``            ``res = res ^ arr[i];``        ``}``        ``return` `res;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };``        ``int` `n = arr.Length;``        ``Console.Write(getOddOccurrence(arr, n));``    ``}``}` `// This code is contributed by Sam007`

## PHP

 ``

## Javascript

 ``

Output :

`5`

Time Complexity: O(n)
Auxiliary Space: O(1)

#### Method 3:Using Built-in Python functions:

• Count the frequencies of every element using the Counter function
• Traverse in frequency dictionary
• Check which element has an odd frequency.
• Print that element and break the loop

Below is the implementation:

## Python3

 `# importing counter from collections``from` `collections ``import` `Counter` `# Python3 implementation to find``# odd frequency element``def` `oddElement(arr, n):` `    ``# Calculating frequencies using Counter``    ``count_map ``=` `Counter(arr)` `    ``for` `i ``in` `range``(``0``, n):` `        ``# If count of element is odd we return``        ``if` `(count_map[arr[i]] ``%` `2` `!``=` `0``):``            ``return` `arr[i]`  `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[``1``, ``1``, ``3``, ``3``, ``5``, ``6``, ``6``]``    ``n ``=` `len``(arr)``    ``print``(oddElement(arr, n))` `# This code is contributed by vikkycirus`

Output:

`5`

Time Complexity: O(N)
Auxiliary Space: O(N)

#### Method 4:Using HashSet

This problem can also be solved using HashSet by traversing the array and inserting element if not already present else deleting the element from the HashSet. So, after the traversal is complete the only element left in the HashSet is the element which is present three times.

## C++

 `// C++ program to find the element``// occurring odd number of times``#include``using` `namespace` `std;` `// Function to find the element``// occurring odd number of times``int` `getOddOccurrence(``int` `arr[], ``int` `N)``{``     ``unordered_set<``int``> s;``     ``for``(``int` `i=0; i

## Java

 `/*package whatever //do not write package name here */` `// Java program to find the element``// occurring odd number of times``import` `java.io.*;``import` `java.util.*;` `class` `GFG {``    ``public` `static` `int` `getOddOccurrence(``int` `arr[], ``int` `N)``    ``{``        ``HashSet s = ``new` `HashSet<>();``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``if` `(s.contains(arr[i])) {``                ``s.remove(arr[i]);``            ``}``            ``else` `{``                ``s.add(arr[i]);``            ``}``        ``}` `        ``int` `ans = ``0``;``        ``for` `(``int` `val : s) {``            ``ans = val;``            ``break``;``        ``}``        ``return` `ans;``    ``}``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``2``, ``3``, ``5``, ``4``, ``5``, ``2``, ``4``, ``3``, ``5``, ``2``, ``4``, ``4``, ``2` `};``        ``int` `n = arr.length;` `        ``// Function calling``        ``System.out.println(getOddOccurrence(arr, n));``    ``}``}`

Output

`5`

Time Complexity: O(N)
Auxiliary Space: O(N)

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